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Context Let $\mathbb{k}$ be an algebrically closed field. Let us consider the additive group $\mathbb{G}_a \simeq \mathbb{k}$ and the multiplicative group $\mathbb{G}_m \simeq \mathbb{k} \backslash \{0\}$.

I wanted to show that these groups are not isomorphic. Here is my argument:

There is no order $2$ elements in $\mathbb{G}_a$. Indeed, if $x$ was a non-zero element of order $2$ then $2x = 0$ then $x=0$ which is contradictory. However, in the other side $-1$ is of order $2$ in $\mathbb{G}_m$.

Is my argument right?

I think that there may be a problem with caracteristic 2 fields. So I'm looking for a better argument that does not have any caracteristic issues.

Thanks in advance.

K. Y.

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    $\begingroup$ It depends if you want to show that there is no isomorphism of abstract group between the $k$-points of the algebraic groups, or no isomorphism of algebraic group. Your argument actually shows there is no abstract group isomorphism if the characteristic is not $2$. $\endgroup$
    – user120527
    Commented Apr 5, 2018 at 11:44
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    $\begingroup$ I think you mean that there is no element of order $2$ in $\mathbb{G}_a$... $\endgroup$
    – Arnaud D.
    Commented Apr 5, 2018 at 11:44
  • $\begingroup$ If the characteristic is $2$, consider elements of order $3$. $\endgroup$
    – Andreas Blass
    Commented Apr 5, 2018 at 13:33
  • $\begingroup$ @user120527 Actually I meant to prove that there is no isomphism as algebraic group but I thought that it would be enough that there is no isomorphism as abstract group. $\endgroup$
    – Kal_Aki
    Commented Apr 5, 2018 at 20:12
  • $\begingroup$ @ArnaudD. You're right. Sorry for the mistake, I'll edit it. $\endgroup$
    – Kal_Aki
    Commented Apr 5, 2018 at 20:12

1 Answer 1

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Your answer is correct when the characteristic is not $2$.

For an answer that works in any characteristic : show that all elements of $\mathbb{G}_a$ have the same (potentially infinite) order, while $\mathbb{G}_m$ has elements of all finite orders.

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